Classical-Mechanics · Unit 8 · Video 4 · Interactive Practice

Tension in a Rope: Why a Heavy Rope Pulls Unevenly

IKey Formulas

Formula	Meaning	Applies to
$T(x) = \lvert F_{R,L}(x)\rvert = \lvert F_{L,R}(x)\rvert$	Tension = size of the cut's action–reaction pair	Any rope
$T = F_A$	Uniform tension, equal to the applied pull	Massless rope
$T(x) = \mu_k m_1 g + \left(m_1 + \dfrac{m_2}{d}\,x\right)A$	Tension a distance $x$ from the block	Massive rope
$F_A - \mu_k m_1 g = (m_1 + m_2)\,A$	Newton's 2nd law on block + whole rope	Cross-check

Key Insight: The tension at a cut has two jobs — beat the block's friction and accelerate everything to its left (the block plus the rope behind the cut). So $T$ is smallest at the block and largest at your hand, where it equals $F_A$ .

IIVisualization 1 — Slide the Cut Along the Rope

At any cut a distance $x$ from the block, the tension equals the pull the two halves exert on each other.

💡 With a massless rope ( $m_2 = 0$ ) every cut gives the same value — the tension is uniform and equals the applied pull $F_A$ .

IIIVisualization 2 — Rope Mass Sets the Climb

The heavier the rope, the more steeply the tension climbs from the block to your hand.

Rope mass (m₂) 1

IVQuiz Questions

Problem 1 · Massless Rope

Given: You drag a block across a rough floor with a massless rope, applying $F_A = 20\text{ N}$ — find the tension at the middle of the rope.

✅ Correct! A massless rope transmits the pull undiminished, so

T = F_A = 20\text{ N}

everywhere.

❌ Close, but friction acts on the block, not inside the rope. With zero rope mass the pull passes straight through:

T = F_A = 20\text{ N}

❌ Not quite. For a massless rope the two ends carry the same tension, and it equals the applied pull.

Show solution

Newton's 2nd law on any slice of the rope has mass $\times$ acceleration on the right side. For a massless rope that term is zero, so the horizontal forces balance and the tension is the same at every cut:

T = F_A = 20\text{ N}

Friction acts on the block, not within the rope, so it never reduces the tension the rope carries.

Problem 2 · Where Is Tension Greatest?

Given: A massive uniform rope tied to a block is pulled so the whole system accelerates — where along the rope is the tension largest?

✅ Correct!

T(x)

grows with

x

, so it peaks at the hand, where

T = F_A

❌ That holds only for a massless rope. With real rope mass the tension must accelerate more rope as the cut moves toward your hand, so it is not uniform.

❌ Not quite. The tension at a cut must accelerate everything to its left, and that grows as the cut nears the hand.

Show solution

The tension a distance $x$ from the block is

T(x) = \mu_k m_1 g + \left(m_1 + \frac{m_2}{d}\,x\right)A,

which increases with $x$ . The farther the cut lies from the block, the more rope sits to its left for the tension to accelerate. So $T$ is smallest at the block ( $x=0$ ) and largest at the hand ( $x=d$ ), where it equals the applied pull $F_A$ .

Problem 3 · Tension at Both Ends

Given: $m_1 = 2\text{ kg}$ , rope $m_2 = 1\text{ kg}$ , length $d = 2\text{ m}$ , $\mu_k = 0.5$ , $g = 10\text{ m/s}^2$ , and the system accelerates at $A = 4\text{ m/s}^2$ .

Tension at the block end, $T(0)$ ?

Tension at the hand end, $T(d)$ ?

✅ Correct!

T

runs from

18\text{ N}

at the block to

22\text{ N}

at the hand — a

4\text{ N}

climb.

❌ That is $T(d)$ , the hand end. At the block the rope behind the cut has no mass, so

T(0) = \mu_k m_1 g + m_1 A

❌ Check $T(0)$ . At the block the tension beats friction and accelerates only the block:

T(0) = \mu_k m_1 g + m_1 A

❌ Check $T(d)$ . At the hand the tension beats friction and accelerates block + whole rope:

T(d) = \mu_k m_1 g + (m_1 + m_2)A

Show solution

The kinetic friction on the block is

f = \mu_k m_1 g = 0.5 \times 2 \times 10 = 10\text{ N}.

Block end ( $x = 0$ , no rope behind the cut):

T(0) = f + m_1 A = 10 + 2(4) = 18\text{ N}.

Hand end ( $x = d$ , the whole rope lies behind the cut):

T(d) = f + (m_1 + m_2)A = 10 + (3)(4) = 22\text{ N},

which also equals the applied pull $F_A$ .

Problem 4 · The Tension Gap

Given: A uniform rope of mass $m_2 = 0.5\text{ kg}$ helps drag a block; the whole system accelerates at $A = 4\text{ m/s}^2$ (take $g = 10\text{ m/s}^2$ ). How much larger is the tension at the pulling end than at the block end?

✅ Excellent! The gap is exactly the force to accelerate the rope:

\Delta T = m_2 A = 2\text{ N}

❌ Close — that is the rope's weight $m_2 g = 5\text{ N}$ . The gap comes from accelerating the rope, not supporting it:

\Delta T = m_2 A

❌ They actually cancel.

\mu_k m_1 g

and

m_1 A

appear at both ends, so they subtract away — only the rope's own mass survives:

\Delta T = m_2 A

❌ Zero is the massless case. A massive accelerating rope always has

\Delta T = m_2 A > 0

Show solution

Subtract the two endpoint tensions:

\Delta T = T(d) - T(0) = \big[\mu_k m_1 g + (m_1 + m_2)A\big] - \big[\mu_k m_1 g + m_1 A\big].

The friction term $\mu_k m_1 g$ and the block-inertia term $m_1 A$ are identical at both ends and cancel, leaving only the rope's own inertia:

\Delta T = m_2 A = 0.5 \times 4 = 2\text{ N}.

Notably, the gap is independent of $m_1$ , $\mu_k$ , and $g$ .

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